A 0.1 M solution of a weak diprotic acid H2A with Ka1 and Ka2; if Ka1 is much larger than Ka2, what pH is expected approximately?

• A. Goverened by Ka2 and concentration
• B. Goverened by Ka1 and concentration
• C. Governed by both Ka1 and Ka2 equally
• D. Determined by Ka2 only

Answer: (B)

The main idea is that in a diprotic acid, the pH is controlled by the stronger dissociation that actually contributes protons to the solution. If Ka1 is much larger than Ka2, the first dissociation H2A ⇌ H+ + HA− dominates the proton balance, while the second step HA− ⇌ H+ + A2− is too weak to add much to [H+]. So the pH behaves like that of a weak monoprotic acid with Ka = Ka1 and the given concentration (0.1 M).

A useful estimate is [H+] ≈ sqrt(Ka1 × C0), which leads to pH ≈ 1/2 [pKa1 − log C0]. With C0 = 0.1 M, log C0 = −1, so pH ≈ 1/2 (pKa1 + 1). This captures why the pH is determined by Ka1 and the overall concentration, not by Ka2.